Let $a(x)=2x^4+7x^3+2x^2+5x-4$, and $b(x)=x^2+3x-1$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Explanation: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{2x^4+7x^3+2x^2+5x-4}{x^2+3x-1}$ : We divide ${x^2}$ into ${2x^4}$ to get ${2x^2}$ : $ \hphantom{1567|144474} {2x^2}\\ {{{x^2}+3x-1}}|\overline{{2x^4}+7x^3+2x^2+5x-\ 4}\\ \hphantom{37...8........|}\llap{-}\underline{(2x^4+ 6x^3 -2x^2)}\\ \hphantom{37|3....998.........}+x^3+4x^2 \\ $ [What did we do here?] Next, we divide ${x^2}$ into ${x^3}$ to get ${+x}$, and continue doing this until we find the quotient: $ \hphantom{1567|166664} {2x^2 \ \ {+ \ \ x}\ +\ 1}\\ {{{x^2}+3x-1}}|\overline{2x^4+7x^3+2x^2+5x-4}\\ \hphantom{37...8........|}\llap{-}\underline{(2x^4+ 6x^3 -2x^2)}\\ \hphantom{37|3....998.........}{+x^3}+4x^2+5x \\ \hphantom{37.......888...8......|}\llap{-}\underline{(x^3+3x^2-x)}\\ \hphantom{37|3...............99999.....}{+x^2\ + 6x\ - 4}\\ \hphantom{3788888888888888888.|}\llap{-}\underline{(x^2\ +\ 3x\ -\ 1)}\\ \hphantom{37|3............77777777777......}{+3x-3}\\ $ [What did we do here?] The process stops here because $x^2+3x-1$ is a polynomial of the second degree and $3x-3$ is a polynomial of the first degree. So it follows that ${r(x)}={3x-3}$, ${q(x)}={2x^2+x+1}$, and $ \dfrac{2x^4+7x^3+2x^2+5x-4}{x^2+3x-1}={2x^2+x+1}+\dfrac{{3x-3}}{x^2+3x-1}$ To conclude, $q(x)=2x^2+x+1$ $r(x)=3x-3$